# Tutor profile: Jeff G.

## Questions

### Subject: Organic Chemistry

Describe the $$^{1}\mathrm{H}$$NMR spectra of $$\mathrm{CH_3CH_2CH_3}$$.

Both $$\mathrm{CH_3}$$ groups are chemically equivalent, so they appear as one signal. The $$\mathrm{CH_2}$$ group is different, so it appears as another signal. So, the $$^{1}\mathrm{H}$$NMR spectra of $$\mathrm{CH_3CH_2CH_3}$$ has two signals. Now for the splitting... To determine the splitting, we use the $$N+1$$ rule which tells us that a signal is split into $$N+1$$ smaller signals if there are $$N$$ neighboring hydrogens. The $$\mathrm{CH_2}$$ group has 6 neighboring hydrogens, so $$N$$ = 6. Thus the $$\mathrm{CH_2}$$ group gets split into 7 smaller signals (called a septet). The $$\mathrm{CH_3}$$ groups have 2 neighboring hydrogens, so $$N$$ = 2. Thus the $$\mathrm{CH_3}$$ groups are split into 3 smaller signals (called a triplet). In summary, we have a two signals, one a triplet and one a septet in the the $$^{1}\mathrm{H}$$NMR spectra of $$\mathrm{CH_3CH_2CH_3}$$.

### Subject: Chemistry

Bromine consists of two isotopes, $$^{79}\mathrm{Br}$$ and $$^{81}\mathrm{Br}$$. The average atomic mass of Bromine is 79.90 amu. Find the abundance of $$^{81}\mathrm{Br}$$.

Whenever we have a question involving isotope abundance, remember to use the equation: (Average Mass) = (Isotope Abundance)(Isotope Mass) + (Isotope Abundance)(Isotope Mass) We are given the average mass of 79.90 amu. We do not know the abundance of $$^{81}\mathrm{Br}$$, so we will call it $$y$$. We know there are only 2 isotopes, so the abundance of $$^{79}\mathrm{Br}$$ must be $$1-y$$. The mass of $$^{79}\mathrm{Br}$$ is 79 and the mass of $$^{81}\mathrm{Br}$$ is 81. Let's put all that into the equation: $$79.9 = (1-y)(79) + (y)(81)$$ Now we can solve for $$y$$: $$79.9 = 79-79y+81y$$ $$0.9=2y$$ $$y = 0.45$$ So the abundance of $$^{81}\mathrm{Br}$$ is 0.45 or 45%.

### Subject: Physics

John Stapp was a pioneer in the field of biophysics--the application of physical principles to the human body. He used a rocket-powered sled to test human tolerance to large accelerations. If the subject is accelerated to a speed of 400 m/s in 1.90 seconds, please find the acceleration. How far does the sled travel during this time?

First, let's write down our given quantities and our unknown quantities: $$v_i = 0$$ m/s $$v_f = 400$$ m/s $$t_i = 0$$ $$t_f = 1.90$$ s $$a = ?$$ $$d = ?$$ To find the acceleration, use the equation $$a=\frac{(v_f - v_i)}{(t_f-t_i)}$$ Now substitute in the known values to find $$a$$: $$a=\frac{(400 \frac{m}{s}- 0)}{(1.90s-0)}=210.5\frac{m}{s^2}$$ To find the distance $$d$$ traveled, let's use the equation: $$d = v_it_f +\frac{1}{2}at_f^2$$ We know $$a$$ since we found it above. Now let's substitute in the values: $$d=0(1.9)+\frac{1}{2}(210.5)(1.9)^2$$ $$d=380$$ m Tip: You could have used the other kinematic equation: $$v_f^2 = v_i^2+2ad$$ to find $$d$$.

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